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Section I.2 Torsion of Non-Circular Bars
The observations made for torsion of members with circular cross sections do not hold for those with non-circular cross sections. Consider the following facts for members with non-circular cross sections:
- The shear stress is not constant at a given distance from the axis of rotation. As a result sections perpendicular to the axis of member warp, indicating out-of-plane displacement.
- The theory of elasticity shows that the shear stress at the corners is zero.
- Maximum shear strain and stress are not at the farthest distance from the rotational axis of a homogeneous non-circular member.
Members with rectangular cross sections:
For a rectangular member under torsion the corners do not distort; the corner square angles remain square after torque is applied. This indicates that shear strain is zero at the corners since there is no distorsion. This fact is illustrated in this figure.
St. Venant was the first to accurately describe the shear stress distribution on the cross section of a non-circular member using the Theory of Elasticity. For more information click here.
Theory of Elasticity shows that:
- the maximum shear strain and stress occur at the centerline of the long sides of the rectangular cross section
- the shear strain and stress at the corners and center of the rectangular cross section are zero.
- the strain and stress variations on the cross section are primarily nonlinear.
- the preceding statements are demonstrated in the following figure.
The theory of Elasticity has been applied to find analytical solutions for the torsion of rectangular elastic members. The resulting equations for shear stress and angle of twist are as follows:
Here is Table A6.1
NOTE: The equations described in this section can only be used for members having OPEN cross-sectional shapes. For open cross sections composed of multiple thin plates refer to Section I.5.
Section I.3 Elastic Membrane Analogy
Prandtl showed that the Laplace equation describing the torsion of an elastic member is identical to that used to describe the deflection of an elastic membrane subjected to a uniform pressure.
The elastic membrane analogy allows the solution of a torsion problem to be determined in a simpler way than that found by the theory of elasticity which requires the availability of the warping function.
Elastic Membrane and Twisted Bar Relationships
Consider a tube which has the same cross-sectional boundary as the bar. For example, if the bar has a solid square cross section of side dimension b, then the tube will have a hollow square cross section of side dimension b as well. Next we stretch an elastic membrane over the tube's cross section and apply internal pressure. The deflected shape of the membrane helps us visualize the stress pattern in the bar under torsion.
The analogy can be viewed as follows:
- Lines of equal deflection on the membrane (contour lines) correspond to shearing stress lines of the twisted bar.
- The direction of a particular shear stress resultant at a point is at right angle to the maximum slope of the membrane at the same point.
- The slope of the deflected membrane at any point, with respect to the edge support plane is proportional in magnitude to the shear stress at the corresponding point on the bar's cross section.
- The applied torsion on the twisted bar is proportional to twice the volume included between the deflected membrane and plane through the supporting edges.
Examples
Keep in mind that the slope at a point on the deflected membrane, and not the displacement from the base, is the parameter that is related to the shear stress in the bar. In all of the following examples, the slope is zero at the very top of the membrane, therefore the stress is zero, not the maximum, at the same location on bar's cross section.
- Figure 1
- Figure 2
- Figure 3 Notice that the stress is maximum at the inside corner and zero at the outside corners.
Section I.4 Torsion of Thin - Wall Open Sections
The sections shown below are extruded sections
EXAMPLE PROBLEMS
- Example Problem 1 Maximum shear stress and angle of twist of a FORMED section
- Example Problem 2 Maximum shear stress and angle of twist of an EXTRUDED section
Angle of twist per unit length
T = Torque
G = Modulus of rigidity
K = Apparent polar moment of inertia (given in table below)
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side = a |  | 
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* This is Table A6.3 in Chapter A6 of "Analysis and Design of Flight Vehicle Structures", by Bruhn.
A more enxtensive list of formulae for various solid and hollow shapes can be found in "Formulas for Stress and Strain", by Roark, 1954 Ed.
I.1-6 Power Transmission
Circular bars or shafts are commonly used for transmission of power. Circular cross section is used because, as explained in the previous section, it will not get distorted under torque. From design stand point, it is necessary for the shaft to be strong enough to transmit power safely without exceeding the shaft material's elastic limit. We recall from physics that for angular motion: Work = Torque x Angular Displacement P = T w where w is the angular velocity of the shaft. This is the relationship between the torque in the shaft and the power transmitted through it from one end to the other. T = HP x 63025 / RPM. Example 1: Transmission of Power Mechanical Coupling Shaft Coupling
Example 2: Mechanical Coupling
Circular bars or shafts are commonly used for transmission of power. Circular cross section is used because, as explained in the previous section, it will not get distorted under torque. From design stand point, it is necessary for the shaft to be strong enough to transmit power safely without exceeding the shaft material's elastic limit. We recall from physics that for angular motion:
Power = d/dt (Work)
If torque is not a function of time, then the equation for power simply becomes:
It is important that we use consistent units for P, T, and w. Power is commonly specified in horsepower, HP. In that case we can convert HP to ft-lb/sec by multiplying it by 550, and to in-lb/sec by multiplying it by 6600. Angular velocity is usually given in revolutions per minute or RPM. It should then be converted to rad/sec. To do this multiply the value in RPM by 2 pi and divide by 60. Therefore, to get the units of in-lb for torque, the conversion factor becomes:
Mechanical coupling is commonly used to extend the length of a shaft or to connect the shaft to another component. An example of this can be found in flap mechanism as well as in the propeller shaft. The figure below shows that with two disks in contact, the applied torque will tend to rotate one relative to the other. This relative displacement, if prevented, will lead to the creation of shear stress and strain at the surface of contact. These types of disks are typically bolted together. In that case, the torque will put the bolts in the state of plane shear.
In problems of mechanical coupling, it is commonly assumed that the surface of contact is frictionless and that the entire torque is transmitted through the bolts which are distributed in a circular pattern at a particular distance from the center.
As in the case of a circular shaft, the shear strain varies linearly from the center of the cross section. Since the diameters of bolts are usually much smaller than that of the coupling plate, it is commonly assumed that the shear strain is constant along the bolt's cross-sectional area. Starting from the original equation for torque we can derive the special form for the mechanical coupling problems as follows:
In shaft coupling, we are concerned with the number and size of bolts to be used in order to keep the stress in the bolts below the allowable value. As a designer we have a choice of materials, size, number, and distribution pattern for the coupling bolts. This freedom allows us to examine different ways that a coupling can be designed such that it is efficient from the stand point of design as well as manufacturability and cost.
We need to balance the number and size of bolts being used. Is it better to use a large quantity of small bolts or few large bolts. We also need to keep in mind the damage tolerance issue. If we use too few bolts, then what would happen in the case one gets sheared off. Will the remaining bolts carry the desired load? Questions like this would have to be answered by the designer before the design can be accepted for production.
Below we see the equations used for the coupling analysis.
Section I.1 Torsion of Circular Bars
Applications:
Rotating Machinery
- Propeller shaft
- Drive shaft
- Landing gear strut
- Flap drive mechanism
When a circular bar is twisted, its cross section remains planeand circular. This characteristic is due to axisymmetric shape of the cross section, and applies to circular bars that are either solid or hollow, homogeneous or non-homogeneous, elastic or inelastic.
Shear Strain Calculation:
Based on the characteristic stated above, the shear strain variation is linear, and is described by the equation
where r is the radial position measured from the center of the cross section and c is the radius of the cross section.
Rule of Thumb:
The torsion-induced shear strain is always a linear function of r with the maximum value at the edge of the cross section. This is true for all possible conditions, whether the bar is elastic or inelastic, homogeneous or non-homogeneous as shown in these examples.
Shear Stress and Angle of Twist Calculations:
Four possible scenarios are considered for shear stress and angle of twist calculations. In each case the equations used for these calculations are explicitly described. It should be obvious that in all of these calculations we are dealing with members having circular cross sections. As you will see the degree of complexity in the analysis grows as we proceed in the following sequence
- I.1-1 Elastic and Homogeneous
- I.1-2 Elastic and Non-homogeneous
- I.1-3 Inelastic and Homogeneous
- I.1-4 Inelastic and Non-homogeneous
- I.1-5 Residual Stress Distribution
- I.1-6 Power Transmission
Restriction: The applied torque(s) must be in a plane(s) perpendicular to the axis of the bar.
The torsion-induced shear stress variation in an elastic, homogeneous, and isotropic bar is determined by
The above shear stress equation is known as the elastic torsion formula, and we call it El-Hocir. This acronym helps us to easily remember under what conditions we can use the elastic torsion formula, viz, elastic, homogeneous, and circular.
The shear stress equation shows that for an elastic bar (i.e., when the maximum shear stress is less than the proportional limit shear stress of the material), the stress varies linearly with radial position. Thus, the maximum shear stress in this case would be at the edge of the cross section (i.e., at the farthest distance from the center). Here are some examples of elastic and inelastic stress variations for homogeneous and non-homogeneous circular bars made of elastoplastic materials.
The angle of twist at one section relative to another can be found in two ways:
In the first equation we are using geometric relationship between angle of twist and shear strain. This is a very powerful equation as it tells us that we don't need to know the torque or the resulting shear stress in order to calculate the relative twist angle. If we know the maximum shear strain at section B and know the distance L between sections A and B, then we can calculate the relative twist angle using this formula. Notice that this equation can also be written in terms of shear strain at any radial position. What is needed is the value of r and its corresponding shear strain.
The second equation is useful when we don't know the shear strain, but we do know the internal torque at section B. Then using this equation we can determine the relative angle of twist.
Total angle of twist at section B is determined by
The linear strain variation is valid as stated before; however, since the bar is non-homogeneous, i.e., made of two or more materials, we cannot use El-Hocir. This problem is statically indeterminate in the sense that we don't know how much of the torque is being carried by each material. The general torque formula in its integral form must be broken up into two or more parts depending on the material make up of the bar. Also, since the bar is totally linearly elastic (i.e., stresses are below the proportional limit), Hooke's law may be used to relate stress to strain in each respective region in the cross section. Here is what the torque equation looks like for a bar made of two materials:
Knowing the maximum shear strain, we can use the above equation to solve for the torque at a given section, or vice versa. Notice that we used the same linear equation for stress-strain relations in both integrals with shear modulus being the only difference. Examining this equation more carefully, we realize that while maximum shear strain always occurs at the farthest distance from the center of the bar, it is not necessary for maximum shear stress to be at the same location. The difference between the shear moduli of the core and the shell could be such that the shear stress in the core at the interface of the shell and core might be higher than that at the edge of the bar near its outside surface.
Rule of Thumb:
Maximum shear strain occurs at r = c regardless of whether the shell material has higher or lower shear rigidity.
Maximum shear stress occurs in the material with the highest shear rigidity regardless of its location. Here are some examples:
Since the core and the shell are assumed to be perfectly attached at the interface, the angle of twist is the same for both materials; therefore,
As mentioned earlier, the shear strain pattern is linear eventhough the member has been stressed beyond the elastic limit and into the inelastic range. For a linearly elastic material, shear stress variation is linear up to the radial distance corresponding to the elastic rim shown in the figures below. Beyond this distance, Hooke's law canNOT be used to relate stress to strain variation. Once again the material is assumed to be elasto-plastic.
Let's see how torsional analysis is performed for a circular bar as the internal torque is increased from zero to that corresponding to the fully-plastic bar. Notice that we are examining only one section (slice) of the bar along its length. Therefore, depending on which section we cosider the internal torque may be different.
If at the section of interest the maximum shear strain is below the elastic-limit shear strain of the material, then the bar is fully elastic and the corresponding torque can be obtained as shown in (1). If the maximum shear strain reaches the elastic shear strain, then the bar has reached the elastic limit state, and the torque calculated according to (2) is referred to as the "elastic-limit torque". As the torque is increased beyond the elastic-limit torque, the internal shear stress moves into the inelastic range. In that case we must first locate the elastic rim by calculating re. Once that distance is known, then we can calculate the internal torque corresponding to the specified max. shear strain according to (3). Note that in this case, the material is assumed to be elasto-plastic, which implies that beyond the elastic limit, stress remains constant at the elastic limit value. If the torque is increased until the cross-section is completely in the inelastic range, then the maximum inelastic or fully-plastic torque is found according to (4). 
The angle of twist at a given section in an inelastic circular bar is found by
Here, we are dealing with a circular bar that is made of two or multiple materials distributed in the radial direction. The bar is twisted beyond the elastic limit of at least one of the materials. It is not necessary for all materials to be stressed beyond their respective elastic limits. In this case, the shear strain variation is still linear and the torque equation is broken up at the material interface and at the point where the transition from elastic to inelastic region takes place.
Let's consider the case where the bar is composed of two different materials. There are several possibilities that need to be investigated: (1) The inner material (core) may become inelastic while the outer shell is still elastic; (2) the outer shell may become fully plastic while the core remains far below its proportional limit; etc.
Here is an example of a non-homogeneous circular bar in a state of inelastic stress and strain. 
It is obvious that the analysis of this type of problem requires more rigor than those dealing with elastic torsion.
Analysis Procedure:
Analysis of circular non-homogeneous bars in which max. shear strain is specified. Notice that this is a statically indeterminate problem as we don't know what percentage of the resultant torque is being carried by each material.
- Knowing that the max. shear strain occurs at the farthest distance from the center of circular cross section, we calculate the actual shear strain at the material interface using the linear equation. Now we know the max. shear strain in each material.
- Next, we calculate the elastic-limit shear strain of each material from the strain-stress diagram or material properties provided in the problem, and compare it with the corresponding actual shear strain. For inelastic torsion problem, the max. shear strain in one of the materials must exceed its corresponding elastic limit value.
- With this information, we can proceed with plotting the stress variation across the cross section of the bar. This gives us a map that we can use to set up the integral form of the torque equation. The integral would be split into several segments depending on the variations of stresses and materials.
- First we consider the core. Based on the stress variation in the core, we may need up to two separate integrals. Then, we examine the shell. Again, depending on the stress variation, we may need up to two separate integrals - one for the elastic part and another for the inelastic part.
- Once we identify the limits on each integral, and correctly write the integrand in each case, we can proceed with the integration to find the torque that corresponds to the specified max. shear strain in the problem statement.
- Notice that the integrals correspond to the portion of the resultant torque carried by the core and the shell. So once we have calculated these values, we can determine the percentage of torque carried by each constituent.
- The angle of twist is calculated in the same manner specified in the previous section. The main assumption is that the two materials are perfectly bonded at the interface.
I.1-5 Residual Stress Distribution in Homogeneous Circular Bars
When a circular bar is torqued beyond its elastic limit, upon the removal of the applied torque, the shear stress may not completely disappear. This remaining stress is known as residual shear stress and is calculated as follows based on the method of superposition. 
In the figure above, the first diagram on the left reflects the elastoplastic behavior of the material during the loading phase (1), and the second diagram in the middle shows the linear behavior of the same material during the unloading phase (in the stress-strain diagram of a linearly elastic material the unloading path is always linear and parallel to the elastic line) (2). By superposing the two stress diagrams, we obtain the residual stress distribution (3). Although there is residual stress without the presence of any external torque, the bar remains in the state of equilibrium as the positive area under the curve equals the negative area. Notice that the three diagrams are not drawn to scale.
If the bar is torqued beyond the elastic limit, there will be a permanent twist upon removal of the torque. This angle of twist is the difference between the angle of twist due to loading and that due to unloading, and is expressed as
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